According to the American Heart Association, \"Cardiomyopathy refers to diseases

Question

According to the American Heart Association, "Cardiomyopathy refers to diseases of the heart muscle. These diseases have many causes, signs and symptoms, and treatments. The heart muscle becomes enlarged, thick or rigid in cardiomyopathy, and in rare cases the muscle tissue is replaced with scar tissue." The following is the left ventricular ejection fraction for a random sample of 17 patients with acute dilated cardiomyopathy:

0.19, 0.24, 0.40, 0.23, 0.30, 0.19, 0.24, 0.32, 0.28, 0.24
0.18, 0.22, 0.14, 0.30, 0.07, 0.12, 0.17

h) Calculate the maximum likelihood estimate for using this data.  

i) Compute a 92% Confidence Interval for .

j)Compute a 92% Prediction Interval for a single future weight measurement.

Explanation / Answer

(h) MLE of =sqrt(sum(Xi-X-)2/(n-1))=sqrt(0.1044)=0.0808

(i) since sample size is less than 30, so we use t-value for finding confidence interval

(1-alpha)*100% confidence interval for population mean=sample mean±t(alpha/2,n-1)*sd/sqrt(n)

92% confidence interval for population mean=mean±t(0.08/2, n-1)*sd/sqrt(n)

(c)

92% prediction interval for sample mean=mean±t(.08/2,4)*sd*sqrt(1+1/n)=0.2253±1.87*0.0808*(1+1/17)=

=0.2253±0.1600=(0.0653,0.3853)

sn x x-mean (x-mean)2 1 0.19 -0.0353 0.0012 2 0.24 0.0147 0.0002 3 0.4 0.1747 0.0305 4 0.23 0.0047 0.0000 5 0.3 0.0747 0.0056 6 0.19 -0.0353 0.0012 7 0.24 0.0147 0.0002 8 0.32 0.0947 0.0090 9 0.28 0.0547 0.0030 10 0.24 0.0147 0.0002 11 0.18 -0.0453 0.0021 12 0.22 -0.0053 0.0000 13 0.14 -0.0853 0.0073 14 0.3 0.0747 0.0056 15 0.07 -0.1553 0.0241 16 0.12 -0.1053 0.0111 17 0.17 -0.0553 0.0031 sum= 3.83 3.6047 0.1044 mean= 0.2253

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